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Let $A$ be an invertible matrix in $M_n(\mathbb{C})$.

How do I prove that $A^{-1}$ has the same block structure in its Jordan canonical form as $A$ does?

For each $x\in \mathbb{C}^n, A(x)=\lambda x$ iff $A^{-1}(x)=1/\lambda x$. Hence, they have exactly the same number of distinct eigenvalues.

So WLOG, we only need to check for the case $A$ has exactly one eigenvalue $\lambda$.

Consider a cycle of generalized eigenvectors of $A$ corresponding to $\lambda$. That is, $\{v_1,...,v_k\}$. However, this does not need to be a cycle of generalized eigenvectors of $A^{-1}$ corresponding to $1/\lambda$. So I don't have a clue here. How do I prove it?

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  • $\begingroup$ Consider an upper-triangular matrix $A=(a_{ij})$ of the form: $a_{ii}=\lambda$, and $a_{i,i+1}\neq0$ for any $i$. What's the Jordan normal form for $A$? $\endgroup$ – Yai0Phah Dec 15 '15 at 21:11
  • $\begingroup$ @FrankScience I don't get your point. That does not ease the problem at all $\endgroup$ – Rubertos Dec 15 '15 at 21:14
  • $\begingroup$ Could you please compute $A^{-1}$ when $A$ is a Jordan block and see how it's like? $\endgroup$ – Yai0Phah Dec 15 '15 at 21:16
  • $\begingroup$ math.stackexchange.com/questions/83323/jordan-canonical-form $\endgroup$ – Rubertos Dec 15 '15 at 21:18
  • $\begingroup$ @FrankScience Here it is. Still I don't get how iy helps.. $\endgroup$ – Rubertos Dec 15 '15 at 21:21
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Suppose the Jordan canonical form of $A$ is given by the block diagonal form $J = \operatorname{diag} (J_1,...,J_m)$, where each $J_k$ has the form $J_k = \begin{bmatrix} \lambda_k & 1 & 0 & \cdots & 0 \\ 0 & \lambda_k & 1 & \cdots & 0 \\ 0 & 0 & \lambda_k & \cdots & 0 \\ \vdots & & & & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_k \end{bmatrix}$. Since $A$ is invertible, we see that each $J_k$ is invertible. Then we see that $J^{-1} = \operatorname{diag} (J_1^{-1},...,J_m^{-1})$.

Hence we only need to establish that the Jordan canonical form of $J_k^{-1}$ has the same form as $J_k$.

One simple way to do this is to note that $J_k v = \lambda_k v_k$ iff $J_k^{-1} v = {1 \over \lambda_k} v$, and so $\ker (J_k -\lambda_k I) = \ker (J_k^{-1}-{1 \over \lambda_k} I)$. In particular, $\dim \ker (J_k^{-1}-{1 \over \lambda_k} I) = 1$ and so the Jordan Canonical form of $J_k^{-1}$ is $\begin{bmatrix} {1 \over \lambda_k} & 1 & 0 & \cdots & 0 \\ 0 & {1 \over \lambda_k} & 1 & \cdots & 0 \\ 0 & 0 & {1 \over \lambda_k} & \cdots & 0 \\ \vdots & & & & \vdots \\ 0 & 0 & 0 & \cdots & {1 \over \lambda_k} \end{bmatrix}$ (with the same dimensions as $J_k$, of course).

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  • $\begingroup$ Hi, nice answer. I am just trying to follow the intuition. Why do we know that dimker(J−1k−1λkI)=1 ? and how does this show that J_k has the same form as J_k^(-1)? $\endgroup$ – ajl123 Oct 11 '18 at 15:42
  • $\begingroup$ @ajl123: A slightly unsatisfying answer: If $J_k^{-1}$ had a different Jordan form then it would have at least two eigenvectors, that is $\dim \ker (J_k^{-1} - {1 \over \lambda_k}I) \ge 2$ which would be a contradiction. $\endgroup$ – copper.hat Oct 12 '18 at 3:40

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